Question: $\dfrac{ 3n + 6p }{ -10 } = \dfrac{ -10n + q }{ -7 }$ Solve for $n$.
Multiply both sides by the left denominator. $\dfrac{ 3n + 6p }{ -{10} } = \dfrac{ -10n + q }{ -7 }$ $-{10} \cdot \dfrac{ 3n + 6p }{ -{10} } = -{10} \cdot \dfrac{ -10n + q }{ -7 }$ $3n + 6p = -{10} \cdot \dfrac { -10n + q }{ -7 }$ Multiply both sides by the right denominator. $3n + 6p = -10 \cdot \dfrac{ -10n + q }{ -{7} }$ $-{7} \cdot \left( 3n + 6p \right) = -{7} \cdot -10 \cdot \dfrac{ -10n + q }{ -{7} }$ $-{7} \cdot \left( 3n + 6p \right) = -10 \cdot \left( -10n + q \right)$ Distribute both sides $-{7} \cdot \left( 3n + 6p \right) = -{10} \cdot \left( -10n + q \right)$ $-{21}n - {42}p = {100}n - {10}q$ Combine $n$ terms on the left. $-{21n} - 42p = {100n} - 10q$ $-{121n} - 42p = -10q$ Move the $p$ term to the right. $-121n - {42p} = -10q$ $-121n = -10q + {42p}$ Isolate $n$ by dividing both sides by its coefficient. $-{121}n = -10q + 42p$ $n = \dfrac{ -10q + 42p }{ -{121} }$ Swap signs so the denominator isn't negative. $n = \dfrac{ {10}q - {42}p }{ {121} }$